Let $$3, 7, 11, 15, \ldots, 403$$ and $$2, 5, 8, 11, \ldots, 404$$ be two arithmetic progressions. Then the sum of the common terms in them is equal to:

AP1 has first term $$a_1 = 3$$ and common difference $$d_1 = 4$$, so its general term is $$3 + 4(n-1) = 4n - 1$$

AP2 has first term $$a_2 = 2$$ and common difference $$d_2 = 3$$, giving the general term $$2 + 3(m-1) = 3m - 1$$. 

Common terms must satisfy $$4n - 1 = 3m - 1$$, i.e., $$4n = 3m$$, thereby forming a new AP with common difference $$\text{lcm}(4, 3) = 12$$.

The first common term is 11 (when $$n=3$$, $$m=4$$), so the common progression is 11, 23, 35, ... with difference 12.

 the last common term not exceeding $$\min(403,404) = 403$$, we solve $$11 + 12(k-1) \le 403$$, yielding $$k-1 \le 32.67$$ and hence $$k = 33$$. 

Thus the last term is $$11 + 12\cdot 32 = 395$$, which belongs to both AP1 ($$395 = 4n - 1 \implies n = 99$$) and AP2 ($$395 = 3m - 1 \implies m = 132$$).

With $$k = 33$$ terms, first term $$a = 11$$, and last term $$l = 395$$, the sum is $$ S = \frac{k}{2}(a + l) = \frac{33}{2}(11 + 395) = \frac{33}{2} \times 406 = 33 \times 203 = 6699 $$. Therefore, the answer is 6699.